When is $f(x) = \frac{ax}{e^{-b}-{(e^{c})}^x}$ monotonic?

Question

Please consider the function $f(x) = \frac{ax}{e^{-b}-{(e^{c})}^x}$ where $(a,b,c)>0$ are positive reals. For what values of $x$ is $f(x)$ monotonic? For example, is this true for $x>0$?

Please note that I made a mistake earlier when specifying $f(x)$. I meant: $f_2(x) = \frac{ax+a}{e^{-b}-{(e^{c})}^x}$ which is perhaps why I'm having difficulty. I will of course accept answers for my originally posted $f(x)$.

Taking the derivative of $f_2(x) = \frac{ax+a}{e^{-b}-{(e^{c})}^x}$ I obtain:

$f'_2=\frac{a}{e^{-b}-e^{cx}}+\frac{ce^{cx}(a+ax)}{(e^{-b}-e^{cx})^2}$

When is $f'_2$ positive?

Answer 1

Using Ittay's hint:

$$f'(x)=a\frac{e^{-b}-e^{cx}+cxe^{cx}}{(e^{-b}-e^{cx})^2}=\frac{a}{(e^{-b}-e^{cx})^2}\left(e^{-b}-(1-c)e^{cx}\right)$$

We must thus find the rightmost factor's sign:

$$e^{-b}-(1-c)e^{cx}\geq 0\Longleftrightarrow e^{-b}\geq (1-c)e^{cx}\ldots$$

Can you take it from here? Using logarithms can be pretty handy here.

Answer 2

Hint. $$f'(x)=\frac{a\exp(b)\left((cx-1)\exp(b+cx)+1\right)}{\left(\exp(b+cx)-1\right)^2}.$$ Your would like to find the extrema, that is each $x$ such that $f'(x)=0$. With that in mind, you can classify each extrema by checking signs as usual.

Answer 3

Since $c>0$, $e^{cx}$ is monotonic increasing, and $e^b-e^{cx}$ is monotonic decreasing. So we get $g(x)=\frac{1}{e^{-b}-e^{cx}}$ is monotonic increasing if $-b<cx$. (BEcause $e^{-b}-e^{cx}$ is positive iff $-b<cx$.)

$f(x)=ax$ is increasing for all $x$ because $a>0$. And if $f$, $g$ are increasing, so $fg$ does. Therefore we get $f(x)g(x)=\frac{ax}{e^{-b}-e^{cx}}$ is monotonous incresing if $x>-b/c$.